.20x^2-5=0

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Solution for .20x^2-5=0 equation: 



.20x^2-5=0

a = .20; b = 0; c = -5;
Δ = b2-4ac
Δ = 02-4·.20·(-5)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$


$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2}{2*.20}=\frac{-2}{0.4}
=-5
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2}{2*.20}=\frac{2}{0.4}
=5
$

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